Question

说我有下表：

+------------+------+--------+
| reviewerID | item | rating |
+------------+------+--------+
|          1 |    1 |       5|
|          1 |    2 |       5|
|          1 |    3 |       5|
|          2 |    4 |       5|
|          2 |    1 |       5|
|          2 |    2 |       5|
+------------+------+--------+

我希望得到评论者1未评级但被评论者2评价的项目，反之亦然。输出应该是这样的：

+------------+------+--------+
| reviewerID | item | rating |
+------------+------+--------+
|          1 |    3 |       5|
|          2 |    4 |       5|
+------------+------+--------+

Answer 1

您可以计算这些项目的评论者数量（在这两位评论者之间），并且只选择一位评论者：

SELECT *
FROM   mytable
WHERE  item IN (SELECT   item
                FROM     mytable
                WHERE    reviewerID IN (1, 2)
                GROUP BY item
                HAVING   COUNT(*) = 1)

Answer 2

这就是你所需要的，以获得理想的结果......

SELECT a.* FROM Reviewer a  
JOIN (  SELECT DISTINCT item FROM Reviewer
        GROUP BY item
        HAVING count(item) < 2) b  
ON a.item = b.item

希望它有所帮助!! 祝你好运!!

用于获取两个用户未评级的项目的SQL查询

2 个答案: